Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.8 AIME Prove it Slovenia

Let $(a_n)$ be a non-constant arithmetic sequence with the initial term $a_1 = 1$ . The terms $a_2$ , $a_5$ , $a_{11}$ form a geometric sequence. Find the sum of the first 2009 terms of the sequence $(a_n)$ .

Solution

Let $d$ be the difference of the sequence $(a_n)$ . Then $a_2 = 1 + d$ , $a_5 = 1 + 4d$ and $a_{11} = 1 + 10d$ . Since $a_2$ , $a_5$ and $a_{11}$ form a geometric progression, we have $(1 + 4d)^2 = (1 + d)(1 + 10d)$ or $6d^2 = 3d$ . Since the sequence is not constant, we conclude that $d = \frac{1}{2}$ and the sum of the first 2009 terms is $2009 + \frac{2009 \cdot 2008}{2} = 1010527$ .

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